Equation for tangent plane at a point on the graph of z=f(x,y)

Find an equation for the tangent plane at the point P (3, 1, 8) on the graph of
z = f(x, y) when f_x(3,1)=1 and f_y(3,1)=4

Since f(1, 8) = 3, the equation of the tangent plane through P is
z = f(3,1) + f_x(3,1) (x - 3) + f_y(3,1) (y - 1)
...= 3 + 1(x - 3) + 4(y - 1).

I hope this helps!

Just in case...

Since f(3, 1) = 8, the equation of the tangent plane through P is
z = f(3,1) + f_x(3,1) (x - 3) + f_y(3,1) (y - 1)
...= 8 + 1(x - 3) + 4(y - 1).

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